Let the polynomial be $f(z)$. One can observe that $$f(z)=\frac{1-z^{15}}{1-z^{5}}+z^{15} \frac{1-z^{15}}{1-z^{3}}=\frac{1-z^{20}}{1-z^{5}}+z^{18} \frac{1-z^{12}}{1-z^{3}}$$ so all primitive 15th roots of unity are roots, along with -1 and $\pm i$. To show that there are no more, we can try to find $\operatorname{gcd}(f(z), f(1 / z))$. One can show that there exist $a, b$ so that $z^{a} f(z)-z^{b} f(1 / z)$ can be either of these four polynomials: $$\begin{aligned} \left(1+z^{5}+z^{10}\right)\left(1-z^{32}\right), & \left(1+z^{5}+z^{10}+z^{15}\right)\left(1-z^{30}\right) \\ \left(1+z^{3}+z^{6}+z^{9}+z^{12}\right)\left(z^{32}-1\right), & \left(1+z^{3}+z^{6}+z^{9}\right)\left(z^{30}-1\right) \end{aligned}$$ Thus any unit circle root of $f(z)$ must divide the four polynomials $\left(1-z^{15}\right)\left(1-z^{32}\right) /\left(1-z^{5}\right)$, $\left(1-z^{20}\right)\left(1-z^{30}\right) /\left(1-z^{5}\right),\left(1-z^{15}\right)\left(1-z^{32}\right) /\left(1-z^{3}\right),\left(1-z^{12}\right)\left(1-z^{30}\right) /\left(1-z^{3}\right)$. This implies that $z$ must be a primitive $k$th root of unity, where $k \in\{1,2,4,15\}$. The case $k=1$ is clearly extraneous, so we are done.